Re: [Qemu-devel] Why qemu write/rw speed is so low?

[Stefan Hajnoczi]

On Fri, Sep 09, 2011 at 04:04:07PM +0200, Kevin Wolf wrote:
> Am 09.09.2011 15:54, schrieb Stefan Hajnoczi:
> > On Fri, Sep 9, 2011 at 2:48 PM, Zhi Yong Wu <address@hidden> wrote:
> >> On Fri, Sep 9, 2011 at 6:38 PM, Stefan Hajnoczi
> >> <address@hidden> wrote:
> >>> On Fri, Sep 09, 2011 at 05:44:36PM +0800, Zhi Yong Wu wrote:
> >>>> Today, i did some basical I/O testing, and suddenly found that qemu 
> >>>> write and rw speed is so low now, my qemu binary is built on commit 
> >>>> 344eecf6995f4a0ad1d887cec922f6806f91a3f8.
> >>>>
> >>>> Do qemu have regression?
> >>>>
> >>>> The testing data is shown as below:
> >>>>
> >>>> 1.) write
> >>>>
> >>>> test: (g=0): rw=write, bs=512-512/512-512, ioengine=libaio, iodepth=1
> >>>
> >>> Please post your QEMU command-line.  If your -drive is using
> >>> cache=writethrough then small writes are slow because they require the
> >>> physical disk to write and then synchronize its write cache.  Typically
> >>> cache=none is a good setting to use for local disks.
> >> Now i can not access my workstation in the office.
> >> -drive if=virtio,cache=none,file=xxxx
> >>
> >>>
> >>> The block size of 512 bytes is too small.  Ext4 uses a 4 KB block size,
> >>> so I think a 512 byte write from the guest could cause a 4 KB
> >>> read-modify-write operation on the host filesystem.
> >> You mean RCU? What is its work procedure? Can you explain in more
> >> details if you are available?
> > 
> > If the host file system manages space in 4 KB blocks, then a 512 byte
> > to an unallocated part of the file causes the file system to find 4 KB
> > of free space for this data.  Since the write is only 512 bytes and
> > does not cover the entire 4 KB region, the file system initializes the
> > remaining 3.5 KB with zeros and writes out the full 4 KB block.
> > 
> > Now if a 512 byte write comes in for an allocated 4 KB block, then we
> > need to read in the existing 4 KB, modify the 512 bytes in place, and
> > write out the 4 KB block again.  This is read-modify-write.  In this
> > worst-case scenario a 512 byte write turns into a 4 KB read followed
> > by a 4 KB write.
> 
> But that should only happen with a 4k sector size, otherwise there's no
> reason for RMW.

You're right.  For cache=none (O_DIRECT), the host file system should
not need to do read-modify-write because it can write the single sector
without caring what is in the surrounding 3.5 KB.

Stefan